# Mathematical Induction by ert554898

VIEWS: 2 PAGES: 5

• pg 1
```									Mathematical Induction
Mathematical induction
• Basis step:       Verify that S(1) is true.

• Inductive step:   Assume S(k) is true.
Prove S(k+1) is
true.

• Conclusion:       Therefore S(n) is true
Cont…
• 1+3+5+…+(2n-1)=n2 for every positive
integer n

Step 1: To prove S(1) is true
(2(1)-1) = 12
1=1
Because LHS = RHS  S(1) is true
• Step 2: Assume S(k) is true 1+3+5+…+(2k-1)=k2
is true ------(i)

• Step 3: Prove S(k+1) is true
1+3+5+…+(2k-1)+(2(k+1)-1)=(k+1)2
From (i) 1+3+5+…+(2k-1)=k2
k2 + (2(k+1)-1) = k2 + 2k+1= (k+1)2
Because LHS = RHS S(k+1) is true
Since we can prove S(1) and S(k+1) are true the
S(n) is true.
Show that 5n -1 is divisible by 4 for all n > 4
– Step 1: if n=1, 51 – 4=4 which is divisible by
4
– Step 2: Assume 5n -1 is divisible by 4 then
we have to 5n+1 -1 is divisible by 4.
• 5n+1 -1= 5n.51 -1= 5n.(4+1)-1= 5n.4+ 5n -1
• 5n.4 is divisible by 4 and 5n -1 is also divisible by 4.
•  5n+1 -1 is divisible by 4
– Step 3: Based on the results of step 1 and
2 5n -1 is divisible by 4

```
To top